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jleach
jleach is offline
#8
Oct7-10, 05:28 PM
P: 17
Before I leave for the weekend, I'll give you a preview of what I can remember. Let's first assign the function F(n) to be the procedure for generating the number of partitions for the number n. Let k be a number equal to or less than n. The number of partitions that contain the value k at least one time is F(n-k). But, that is old news. If S is a set of values, we can also let k be the sum of those values within the set. F(n-k) will tell me how many partitions of n contain the set at least once. Now for the number of partitions of n that contain no repeated values, I must double check my notes to be certain, but I know that k is calculated from the sum of natural numbers. I'll try to confirm and give you details next week.